Gradient functions - Vector quantity, which is associated with the definition of private derived functions. The direction of the gradient indicates the path of the formal growth of the function from one point of the scalar field to other.

Instruction

1. To solve the problem of the function, the methods of differential calculus are used, namely, finding partial derivatives of the first order in three variables. It is assumed that the function itself and all its private derivatives own the property of continuity in the field of function definition.

2. The gradient is a vector, the direction of which indicates the direction of the maximum rapid increase in the function F. For this, two points M0 and M1 are selected on the graph, which are the ends of the vector. The magnitude of the gradient is equal to the speed of increasing the function from the point M0 to the point M1.

3. The function is differentiable in all points of this vector, in effect, the projections of the vector on the coordinate axes are all its private derivatives. Then the gradient formula looks further: grad \u003d (? F /? X) i + (? F /? Y) j + (? F /? Z) k, where i, j, k is the coordinates of a single vector. In other words, the gradient of the function is a vector whose coordinates are its private derivatives GRAD F \u003d (? F /? X ,? F /? Y ,? F /? Z).

4. Example1. The function f \u003d sin (x z?) / Y is set to specified. It is required to detect its gradient at the point (? / 6, 1/4, 1).

5. Decision. Use private derivatives for any variable: F'_h \u003d 1 / y computers (x z?) Z?; F'_y \u003d sin (x z?) (-1) 1 / (y?); F'_Z \u003d 1 / y computers (x z?) 2 x z.

6. Submire the famous points of the point coordinate: F'_x \u003d 4 computers (? / 6) \u003d 2? 3; F'_Y \u003d sin (? / 6) (-1) 16 \u003d -8; F'_Z \u003d 4 computers (? / 6) 2? / 6 \u003d 2? /? 3.

7. Apply the function gradient formula: GRAD F \u003d 2? 3 i - 8 j + 2? /? 3 K.

8. EXAMPLE 25. BUY COOLS OF THE GRADIENT OF THE FUNCTION FU \u003d Y ARSTG (Z / X) at point (1, 2, 1).

9. Solution .F'_h \u003d 0 ARSTG (z / x) + y (ARSTG (z / x)) '_ x \u003d y 1 / (1 + (z / x)?) (-Z / x?) \u003d -Yz / (x? (1 + (z / x)?)) \u003d -1; f'_y \u003d 1 arstg (z / x) \u003d arstg 1 \u003d? / 4; F'_Z \u003d 0 ARSTG (z / x) + y (ARSTG (z / x)) '_ z \u003d y 1 / (1 + (z / x)?) 1 / x \u003d y / (x (1 + (z / x)?)) \u003d 1.GRAD \u003d (- 1,? / 4, 1).

The gradient of the scalar field is a vector magnitude. Thus, it is necessary to determine all the components of the corresponding vector, based on the knowledge of the division of the scalar field.

Instruction

1. Read in the textbook on the highest mathematics, which represents the gradient of the scalar field. As a peculiar, this vector value has a direction characterized by the maximum downtime of the scalar function. Such a sense of this vector value is justified by the expression to determine its component.

2. Remember that any vector is determined by its component. The components of the vector are actually projections of this vector on one or other coordinate axis. Thus, if a three-dimensional space is considered, then the vector should have three components.

3. Write down how the components of the vector, which is the gradient of some field are determined. All of the coordinates of such a vector is equal to the derivative of the scalar potential in a variable whose coordinate is calculated. That is, if you need to calculate the "oscus" component of the field of the gradient of the field, it is necessary to prejudice the scalar function along the "X" variable. Please note that the derivative should be private. This means that when differentiating, the other variables that do not participate in it should be considered constants.

4. Write an expression for a scalar field. As it is famous, this term implies every only scalar function of several variables that are also scalar values. The number of variables of the scalar function is limited to the size of the space.

5. Differentiate separately scalar function for any variable. As a result, you will have three new features. Enter all the function in the expression for the gradient vector of the scalar field. Each of the functions obtained is actually an indicator with a single vector of this coordinate. Thus, the gradient final vector should look like a polynomial with indicators in the form of derived functions.

When considering issues involving the representation of the gradient, each function is perceived as scalar fields. It is necessary to introduce the appropriate designations.

You will need

• - Buman;
• - a pen.

Instruction

1. Let the function set the three arguments u \u003d f (x, y, z). A private derivative function, for example, is defined as a derivative of this arion obtained when fixing the rest of the arguments. For the rest of the arguments like. The designations of the private derivative is written in the form: df / dh \u003d u'x ...

2. The full differential will be equal to DU \u003d (DF / DX) DX + (DF / D) DY + (DF / DZ) DZ. Included derivatives are permitted to understand how derivatives in the directions of coordinate axes. The question of finding a derivative in the direction of the specified vector S at the point M (x, y, z) appears at the point M (x, y, z) (do not forget that the direction S specifies the unit vector-ort S ^ O). At the same time, the vector-differential arguments (DX, DY, DZ) \u003d (DSCOS (ALPHA), DSSOs (beta), DSOs (gamma)).

3. Considering the type of complete DU differential, it is allowed to draw the result that the derivative in the direction S at the point M is equal to: (DU / DS) | M \u003d ((DF / DX) | M) COs (alpha) + ((df / d) | M) COs (beta) + ((df / dz) | m) COs (gamma). If S \u003d S (SX, SY, SZ), then guide cosines (computers (alpha), computers (beta), computers ( Gamma)) are calculated (see Fig. 1a).

4. Determining the derivative in the direction, counting point M of variable, allowed to rewrite in the form of a scalar product: (DU / DS) \u003d ((df / dh, df / df / dz), (computers (alpha), computers (beta), comp (Gamma))) \u003d (GRAD U, S ^ O). This expression will be objectively for the scalar field. If it is considered easy to function, then Gradf is a vector having coordinates coinciding with private derivatives F (x, y, z) .Gradf (x, y, z) \u003d ((df / dh, df / df / df / dz) \u003d) \u003d (df / dh) i + (df / d) j + (df / dz) k. Here (i, j, k) are orts of coordinate axes in a rectangular Cartesian coordinate system.

5. If you apply the differential vector-operator Hamilton recruit, then Gradf is allowed to write down how multiplying this vector operator on a scalar f (see Fig. 1b). From the point of view of communication GRADF with a derivative in the direction, equality (GRADF, S ^ O) \u003d 0 is permissible if these vectors are orthogonal. Consequently, Gradf is often determined as the direction of the fastest metamorphosis of the scalar field. And from the point of view of differential operations (GRADF is one of them), Gradf properties exactly repeat the properties of differentiation of functions. In particular, if F \u003d UV, then GRADF \u003d (Vgradu + U Gradv).

Video on the topic

Gradient This is a tool, in graphic editors, performing the fill silhouette with a smooth transition of the same color in other. Gradient It may give a silhouette the result of volume, imitate lighting, glare light on the surface of the subject or the result of the sunset in the background of the photo. This tool has widespread use, in effect, to process photos or creating illustrations, the somewhat will significantly study them to use it.

You will need

• Computer, Adobe Photoshop graphic editor, Corel Draw, Paint.Net or other.

Instruction

1. Open the image in the program or make a new one. Make a silhouette or highlight the outfit area on the image.

2. Turn on the gradient tool on the Graphic Editor toolbar. Place the mouse cursor to the point within the selected area or the silhouette in which the 1st color of the gradient began. Press and hold the left key of the mouse. Move the cursor to the point in which the gradient should go to the final color. Release the left mouse button. The highlighted silhouette will fill the fill with a gradient.

3. Gradient It is allowed to set transparency, colors and their ratio at a certain point of fill. To do this, open the edit window of the gradient. In order to open the editing window in Photoshop - click on the example of the gradient in the "Parameters" panel.

4. In the window that opens, the available options for gradient fill are displayed in the form of examples. In order to edit one of the options, select its mouse click.

5. At the bottom of the window, an example of a gradient is displayed as a wide scale on which the sliders are located. The sliders denote the points in which the gradient should have specified calls, and in the range between the slider, the color evenly moves from the 2nd point color set in the first point.

6. The sliders, which are located at the top of the scale set the transparency of the gradient. In order to change transparency, click on the required slider. The field will appear under the scale in which enter the required degree of transparency in percent.

7. The sliders at the bottom of the scale set the colors of the gradient. Changing one of them, you will be able to prefer the surf color.

8. Gradient May have several transition colors. In order to set another color - click on the free place at the bottom of the scale. It will appear another slider. Set the desired color for it. The scale will display an example of a gradient with another point. You can move the slider by holding them with the support of the left mouse button, in order to achieve the desired combination.

9. Gradient There are several types that can give the shape flat silhouettes. Let's say in order to give a circle the shape of the ball applies a radial gradient, and in order to give the shape of the cone - cone-shaped. In order to give the surface, the illusion of the bulge is allowed to take advantage of the mirror gradient, and the diamond gradient can be used to create glare.

Video on the topic

Video on the topic

Brief theory

The gradient is called the vector, the direction of which indicates the direction of the maximum rapid increase in the function f (x). Finding this vector value is associated with the definition of private derived functions. The derivative in the direction is a scalar value and shows the rate of change of function when driving along the direction specified by some vector.

An example of solving the problem

The task

Danies feature, point and vector. To find:

The solution of the problem

Finding a gradient function

1) We will find the gradient of the function at the point:

The desired gradient:

Finding a derivative in the direction of the vector

2) Find a derivative in the direction of the vector:

where-China, formed by vector and axis

The desired derivative at the point:

The price strongly affects the urgency of the solution (from day to several hours). Online assistance on the exam / standings is carried out by appointment.

The application can be left directly in the chat, having previously throwing the condition of tasks and informing the decision you need. Answer time - a few minutes.

From the school course of mathematics it is known that the vector on the plane is a directed segment. His beginning and end have two coordinates. The coordinates of the vector are calculated by subtracting from the coordinates of the end of the coordinates of the beginning.

The concept of a vector can also be distributed to N-dimensional space (instead of two coordinates there will benchordinates).

Gradientgradz Functionsz \u003d f (x 1, x 2, ... x n) is called the vector of private derived functions at the point, i.e. Vector with coordinates.

It can be proved that the gradient of the function characterizes the direction of the formal growth of the function level at the point.

For example, for the function z \u003d 2x 1 + x 2 (see Figure 5.8), the gradient will at any point will have coordinates (2; 1). It is possible to build it on the plane in various ways by taking any point as the beginning of the vector. For example, you can connect a point (0; 0) with a point (2; 1), or point (1; 0) with a point (3; 1), or point (0; 3) with a point (2; 4), or t .P. (See Figure 5.8). All vectors constructed in this way will have coordinates (2 - 0; 1 - 0) \u003d (3 - 1; 1 - 0) \u003d (2 - 0; 4 - 3) \u003d (2; 1).

Figure 5.8 is clearly seen that the level of function is growing in the direction of the gradient, since the constructed level lines correspond to the values \u200b\u200bof the level 4\u003e 3\u003e 2.

Figure 5.8 - Gradient function z \u003d 2x 1 + x 2

Consider another example - the function z \u003d 1 / (x 1 x 2). The gradient of this function will not be equally the same in different points, since its coordinates are determined by formulas (-1 / (x 1 2 x 2); -1 / (x 1 x 2 2))).

Figure 5.9 shows the function levels of the function bel \u003d 1 / (x 1 x 2) for levels 2 and 10 (direct 1 / (x 1 x 2) \u003d 2 is indicated by dotted line, and the straight line 1 / (x 1 x 2) \u003d 10 - solid line).

Figure 5.9 - Gradients of the function z \u003d 1 / (x 1 x 2) at different points

Take, for example, the point (0.5; 1) and calculate the gradient at this point: (-1 / (0.5 2 * 1); -1 / (0.5 * 1 2)) \u003d (-4; - 2). Note that the point (0.5; 1) lies on the level line 1 / (x 1 x 2) \u003d 2, forz \u003d f (0.5; 1) \u003d 1 / (0.5 * 1) \u003d 2. To Pail with a vector (-4; -2) in Figure 5.9, connect the point (0.5; 1) with a point (-3,5; -1), for (-3,5 - 0.5; -1 - 1) \u003d (-4; -2).

Take another point on the same level line, for example, point (1; 0.5) (z \u003d f (1; 0.5) \u003d 1 / (0.5 * 1) \u003d 2). Calculate the gradient at this point (-1 / (1 2 * 0.5); -1 / (1 * 0.5 2)) \u003d (-2; -4). To portray it in Figure 5.9, connect the point (1; 0.5) with a point (-1; -3.5), for (-1 - 1; -3.5 - 0.5) \u003d (-2; - four).

Take another point on the same level line, but only now in an inseparable coordinate quarter. For example, the point (-0.5; -1) (z \u003d f (-0.5; -1) \u003d 1 / ((- 1) * (- 0.5)) \u003d 2). The gradient at this point will be equal to (-1 / ((- 0.5) 2 * (- 1)); -1 / ((- 0.5) * (- 1) 2)) \u003d (4; 2). Show it in Figure 5.9, connecting the point (-0.5; -1) with a point (3.5; 1), for (3.5 - (-0.5); 1 - (-1)) \u003d (4 ; 2).

It should be noted that in all three cases considered, the gradient shows the direction of growth level growth (towards the line 1 / (x 1 x 2) \u003d 10\u003e 2).

It can be proved that the gradient is always perpendicular to the level line (level surface) passing through this point.

Extremmas functions of many variables

We define the concept extremumfor the function of many variables.

The function of many variables f (x) has at point x (0) maximum (minimum),if there is such a neighborhood of this point, that for all points x from this neighborhood, inequalities are performed (x) f (x (0)) ().

If these inequalities are performed, as strict, then the extremum is called strong, and if not, then weak.

Note that the extremum defined in this way wears localintegrity, since these inequalities are performed only for some neighborhood of the extremum point.

We have a necessary condition for the local extremum of the differentiable function z \u003d f (x 1,.., X n) at the point is the equality of zero of all private derivatives of the first order at this point:
.

Points in which these equality are performed are called stationary.

In a different way, the necessary extremum condition can be formulated as follows: At the point of extremum, the gradient is zero. It is possible to prove more general approval - at the extremma point they turn into zero derivatives in all directions.

Stationary points should be subjected to additional studies - there are sufficient conditions for the existence of a local extremum. For this, the sign of the second order differential is determined. If with any, not equal at the same time zero, it is always negative (positive), then the function has a maximum (minimum). If it may apply to zero not only at zero increments, the question of extremma remains open. If you can take both positive and negative values, then there is no extremum in a stationary point.

In general, the definition of the differential sign is a rather complicated problem, which we will not consider here. For the function of two variables, you can prove that if in a stationary point
The extremum is present. In this case, the sign of the second differential coincides with the sign
. if a
, then this is the maximum, and if
, then this is at least. If a
then there is no extremum at this point, and if
The question of extremma remains open.

Example 1.. Find Extreme Functions
.

We find private derivatives by logarithmic differentiation.

ln z \u003d ln 2 + ln (x + y) + ln (1 + xy) - ln (1 + x 2) - ln (1 + y 2)

Similarly
.

Find stationary points from the system of equations:

Thus, four stationary points were found (1; 1), (1; -1), (-1; 1) and (-1; -1).

We find private derivatives of the second order:

ln (z x `) \u003d ln 2 + ln (1 - x 2) -2ln (1 + x 2)

Similarly
;
.

As
, sign of expression
only OT depends
. Note that in both of these derivatives, the denominator is always positive, so you can consider only the numerator sign, or even the sign of expressions x (x 2 - 3) and (Y 2 - 3). We define it in each critical point and check the execution of sufficient conditions of extremum.

For a point (1; 1) we get 1 * (1 2 - 3) \u003d -2< 0. Т.к. произведение двух отрицательных чисел
\u003e 0, and
< 0, в точке (1; 1) можно найти максимум. Он равен
= 2*(1 + 1)*(1 +1*1)/((1 +1 2)*(1 +1 2)) = = 8/4 = 2.

For a point (1; -1) we get 1 * (1 2 - 3) \u003d -2< 0 и (-1)*((-1) 2 – 3) = 2 > 0. Because The work of these numbers
< 0, в этой точке экстремума нет. Аналогично можно показать, что нет экстремума в точке (-1; 1).

For a point (-1; -1) we obtain (-1) * ((- 1) 2 - 3) \u003d 2\u003e 0. Because Work of two positive numbers
\u003e 0, and
\u003e 0, at the point (-1; -1) you can find a minimum. It is 2 * ((- 1) + (-1)) * (1 + (- 1) * (- 1)) / ((1 + (- 1) 2) * (1 + (- 1) 2) ) \u003d -8/4 \u003d -2.

To find globalthe maximum or minimum (the largest or smallest value of the function) is somewhat more complicated than the local extremum, since these values \u200b\u200bcan be achieved not only in stationary points, but also at the boundary of the definition area. Explore the behavior of the function on the border of this area is not always easy.

Gradient Function and \u003d f (x, y, z) defined in some region. Space (X y z), there is vector With projections indicated by symbols: GRAD Where i, J, K - Coordinate orthops. G. F. - there is a point function (x, y, z), i.e. it forms a vector field. Derivative in the direction of the city of f. At this point reaches the greatest value and equal to: The direction of the gradient is the direction of the most ascent increase of the function. G. F. At this point, perpendicular to the surface surface passing through this point. The effectiveness of the use of G. F. In lithological studies, it was shown in the study of the eol. Central doodles.

Geological Dictionary: in 2 volumes. - M.: Nedra. Edited by K. N. Paffengolts and others.. 1978 .

Watch what is a "quality gradient" in other dictionaries:

This article is about mathematical characteristics; For the method of fill, see: Gradient (Computer Graphics) ... Wikipedia

- (Lat.). The difference in barometric and thermometric indications in different locations. A dictionary of foreign words included in the Russian language. Chudinov A.N., 1910. Gradient difference in the testimony of the barometer and the thermometer in the same moment ... ... Dictionary of foreign words of the Russian language

gradient - Change the value of a certain amount per unit distance in a given direction. The topographic gradient is a change in the height of the terrain on the measured horizontal distance. Themes Relay Protection EN Gradient of The Differential Protection Tripping Characteristic ... Technical translator directory

Gradient - The vector directed towards the apparent increase in the function and equal to its derivative in this direction: where the symbols Ei indicate the unit vectors of the coordinate axes (orts) ... Economics and Mathematical Dictionary

One of the basic concepts of vector analysis and the theory of nonlinear mappings. The gradient of the scalar function of the vector argument from the Euclidean space E n is. The derivative function f (t). According to the vector argument T, that is, N dimensional vector with ... ... Mathematical encyclopedia

Gradient physiological - - the value reflecting the change to either the function indicator depending on the other value; For example, the gradient of partial pressure is the difference in partial pressure, which determines the diffusion of gases from the alveoli (axes) into the blood and from the blood in ... ... Dictionary of terms in physiology of farm animals

I gradient (from lat. Gradiens, born. Paddle gradient grading) vector showing the direction of the separated change of some value, the value of which varies from one point of space to another (see the theory fields). If the magnitude ... ... Great Soviet Encyclopedia

Gradient - (from lat. Gradiens walking, coming) (in mathematics) vector showing the direction of the definite increase in some function; (in physics) measure of increasing or decrease in space or on the plane of any physical size per unit ... ... The start of modern natural science

Books

• Methods for solving some tasks of selected sections of higher mathematics. Workshop, Klimenko Konstantin Grigorievich, Levitskaya Galina Vasilyevna, Kozlovsky Evgeny Aleksandrovich. This workshop discusses the methods of solving some types of tasks from such sections of the generally accepted course of mathematical analysis, as the limit and extreme function, gradient and derivative ...

From the school course of mathematics it is known that the vector on the plane is a directed segment. His beginning and end have two coordinates. The coordinates of the vector are calculated by subtracting from the coordinates of the end of the coordinates of the beginning.

The concept of a vector can also be distributed to N-dimensional space (instead of two coordinates there will be n coordinates).

Gradient GRAD Z functions z \u003d f (x 1, x 2, ... x n) called the vector of private derived functions at the point, i.e. Vector with coordinates.

It can be proved that the gradient of the function characterizes the direction of the formal growth of the function level at the point.

For example, for the function z \u003d 2x 1 + x 2 (see Figure 5.8), the gradient will at any point will have coordinates (2; 1). It is possible to build it on the plane in various ways by taking any point as the beginning of the vector. For example, you can connect a point (0; 0) with a point (2; 1), or point (1; 0) with a point (3; 1), or point (0; 3) with a point (2; 4), or t .P. (See Figure 5.8). All vector constructed vectors will have coordinates (2 - 0; 1 - 0) \u003d
= (3 – 1; 1 – 0) = (2 – 0; 4 – 3) = (2; 1).

Figure 5.8 is clearly seen that the level of function is growing in the direction of the gradient, since the constructed level lines correspond to the values \u200b\u200bof the level 4\u003e 3\u003e 2.

Figure 5.8 - Gradient function z \u003d 2x 1 + x 2

Consider another example - the function z \u003d 1 / (x 1 x 2). The gradient of this function will not be equally the same in different points, since its coordinates are determined by formulas (-1 / (x 1 2 x 2); -1 / (x 1 x 2 2))).

Figure 5.9 shows the level of the level of the function z \u003d 1 / (x 1 x 2) for levels 2 and 10 (direct 1 / (x 1 x 2) \u003d 2 is indicated by dotted line, and the straight
1 / (x 1 x 2) \u003d 10 - solid line).

Figure 5.9 - Gradients of the function z \u003d 1 / (x 1 x 2) at different points

Take, for example, the point (0.5; 1) and calculate the gradient at this point: (-1 / (0.5 2 * 1); -1 / (0.5 * 1 2)) \u003d (-4; - 2). Note that the point (0.5; 1) lies on the line 1 / (x 1 x 2) \u003d 2, for z \u003d f (0.5; 1) \u003d 1 / (0.5 * 1) \u003d 2. To portray the vector (-4; -2) in Figure 5.9, connect the point (0.5; 1) with a point (-3.5; -1), for
(-3,5 – 0,5; -1 - 1) = (-4; -2).

Take another point on the same level line, for example, point (1; 0.5) (z \u003d f (1; 0.5) \u003d 1 / (0.5 * 1) \u003d 2). Calculate the gradient at this point
(-1 / (1 2 * 0.5); -1 / (1 * 0.5 2)) \u003d (-2; -4). To portray it in Figure 5.9, connect the point (1; 0.5) with a point (-1; -3.5), for (-1 - 1; -3.5 - 0.5) \u003d (-2; - four).

Take another point on the same level line, but only now in an inseparable coordinate quarter. For example, the point (-0.5; -1) (z \u003d f (-0.5; -1) \u003d 1 / ((- 1) * (- 0.5)) \u003d 2). The gradient at this point will be equal
(-1 / ((- 0.5) 2 * (- 1)); -1 / ((- 0.5) * (- 1) 2)) \u003d (4; 2). Show it in Figure 5.9, connecting the point (-0.5; -1) with a point (3.5; 1), for (3.5 - (-0.5); 1 - (-1)) \u003d (4 ; 2).